I want to check convergence of sequence
$$y_n = \left(1 - \frac{1}{2^n}, \frac{1}{2^{n+1}}, \frac{1}{2^{n+2}},\dots,\frac{1}{2^{2n}},0,0,0,\dots\right)$$
in $l^1$.
I have some doubts if it's so easy as I think. Does it just converge to $(1,0,0,0,\dots,0) \in \ell^1$?
Or there are some other formal things that I have to show?
EDIT
As suggested in comments I'm applying norm to $y_n - y$, where $y = (1,0,0,\dots,0)$
$$\begin{align} \|y_n-y\| &= \left\| \left(- \frac {1} {2^n}, \frac{1}{2^{n+1}},\dots,\frac{1}{2^{2n}},0, \dots \right) \right\| \\ &= \frac{1}{2^{n}} + \frac{1}{2^{n+1}}+\dots+\frac{1}{2^{2n}} \end{align}$$
and out of these fact I already have the convergence? Because now if $n \rightarrow \infty \Rightarrow \|y_n - y\| \rightarrow0$
Since $$\|y_n-y\|=\frac{1}{2^n}+\dots+\frac{1}{2^{2n}}$$ is expressed as a sum of $n$ terms, each of which goes to zero while $n\to\infty$, it is not "obvious" that the entire sum goes to zero (because while $n$ becomes large we are adding more terms). In order to make sure that this goes to zero, you have to express this sum as a "close form" function of $n$ and then make sure this goes to zero. A counter example: what if we have $\frac{1}{n}+\dots+\frac{1}{n}$ where we are adding $n$ terms? yes, each of those terms goes to $0$ as $n\to\infty$, but the sum is $n$ times $\frac{1}{n}$, so the sum is equal to $1$ and this does not go to $0$ while $n\to\infty$. In our situation though things will work out: In particular, $$\frac{1}{2^n}+\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\dots+\frac{1}{2^{2n}}=\frac{1}{2^n}(1+\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n})=\frac{1}{2^n}\cdot\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}=$$ $$=\frac{1}{2^n}\cdot\frac{2^{n+1}-1}{2^n}=\frac{2^{n+1}-1}{4^n}\to0.$$ So yes, $\|y_n-y\|\to0$.